Problem: $\lim_{x\to -5}\dfrac{10x^2+50x}{x^2-25}=$
Solution: Substituting $x=-5$ into $\dfrac{10x^2+50x}{x^2-25}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{10x^2+50x}{x^2-25}$ can be simplified as $\dfrac{10x}{x-5}$, for $x\neq -5$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-5$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{10x^2+50x}{x^2-25}=\dfrac{10x}{x-5}$ for all $x$ -values in the interval $(-6,-4)$ except for $x=-5$. Therefore, $\lim_{x\to -5}\dfrac{10x^2+50x}{x^2-25}=\lim_{x\to -5}\dfrac{10x}{x-5}=5$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -5}\dfrac{10x^2+50x}{x^2-25}=5$.